#pragma GCC optimize(2)
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <vector>

using namespace std;
using LL = long long;

/*
题意：对于将一个数列重排，使得任意两个不同位值的  i - ai != j - aj

solution: reverse the array  suppose that i < j
    because ai >= aj where i < j  
    so i - ai < j - aj

*/

int n;

void solve(){
    vector<int> nums;
    cin >> n;

    for(int i = 1, x; i <= n; i ++) cin >> x, nums.push_back(x);

    sort(nums.begin(), nums.end());
    reverse(nums.begin(), nums.end());

    for(int i = 0; i < n; i ++){
        cout << nums[i] << " ";
    }
    cout << '\n';
}

int main(){
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    cin >> T;
    while(T--){
        solve();
    }
    return 0;
}